Difficulty: Intermediate to Advanced. Prerequisites: all differentiation rules from Part 2, comfort with algebra.
Once you can differentiate, the real work begins: using derivatives to find tangent lines, analyse business costs and revenues, sketch curves, and optimise. This set of notes also covers basic definite integrals (the reverse of differentiation) and exponential models for compound interest and decay, both of which appear heavily on the exam.
Tangent line
A straight line that touches a curve at exactly one point and has the same slope as the curve at that point. Its equation is found using slope-intercept form: y - y1 = m(x - x1), where m = f'(x1). Think of it as: the best straight-line approximation of the curve at a single point.
Marginal analysis
Using derivatives to approximate the change in cost, revenue, or profit from producing one additional unit. The marginal cost is C'(x), marginal revenue is R'(x), and marginal profit is P'(x). In simple terms: "what does one more unit cost/earn/profit?"
Revenue function
R(x) = price per unit times quantity = p(x) * x. Revenue is what the company brings in from sales.
Profit function
P(x) = R(x) - C(x). Profit is revenue minus total cost.
Average profit function
P-bar(x) = P(x) / x. The profit earned per unit on average.
Critical number
A value of x where f'(x) = 0 or f'(x) is undefined. Critical numbers are the candidates for local maxima and minima. Think of it as: where the slope of the curve is flat (zero) or has a sharp corner.
Local extrema (local max / local min)
A local maximum is a point where f changes from increasing to decreasing. A local minimum is where f changes from decreasing to increasing. Found using the first or second derivative test.
Second derivative test
If f'(c) = 0 and f''(c) > 0, then c is a local minimum. If f'(c) = 0 and f''(c) < 0, then c is a local maximum. If f''(c) = 0, the test is inconclusive. In simple terms: positive second derivative means the curve is concave up (bowl-shaped), so you are at the bottom. Negative means concave down (hill-shaped), so you are at the top.
Inflection point
A point where the concavity of the graph changes (from concave up to concave down, or vice versa). Found where f''(x) = 0 and the concavity actually switches.
Definite integral
The integral from a to b of f(x) dx, written as the integral sign with limits a (bottom) and b (top). It gives the net area between the curve and the x-axis over [a, b]. Evaluated using the Fundamental Theorem of Calculus: F(b) - F(a), where F is an antiderivative of f.
Compound interest (periodic)
A = P(1 + r/n)^(nt), where P is the principal, r the annual rate, n the number of compounding periods per year, and t the time in years.
Continuous compounding
A = Pe^(rt). Used when interest compounds "continuously" (infinitely many times per year).
Exponential decay
A model of the form Q = Q0 * e^(rt) where r < 0. The quantity shrinks over time. The rate of decay r is found by solving for r given two data points.
To find the tangent line at x = a: (1) compute y = f(a) to get the point, (2) compute m = f'(a) to get the slope, (3) plug into point-slope form y - y1 = m(x - x1) and convert to slope-intercept form y = mx + b.
Q26: y = 2x^3 - 3x^2 - 12x + 4 at x = 1. y(1) = 2 - 3 - 12 + 4 = -9. y' = 6x^2 - 6x - 12, so y'(1) = 6 - 6 - 12 = -12. Tangent line: y - (-9) = -12(x - 1), which gives y = -12x + 3.
Horizontal tangent lines occur where f'(x) = 0. For Q26b, solve 6x^2 - 6x - 12 = 0, or 6(x^2 - x - 2) = 0, giving x = 2 and x = -1.
Revenue function: R(x) = p(x) * x, where p(x) is the price/demand function.
Q36: price p(x) = 300 - 2x, so R(x) = (300 - 2x) * x = 300x - 2x^2.
Profit function: P(x) = R(x) - C(x).
Q36: P(x) = (300x - 2x^2) - (40x + 4000) = 260x - 2x^2 - 4000.
Marginal profit: P'(x) = R'(x) - C'(x).
Q36: P'(x) = 260 - 4x.
Average profit: P-bar(x) = P(x)/x = (260x - 2x^2 - 4000)/x = 260 - 2x - 4000/x.
Interpreting P(30) and P'(30) (Q29): P(30) = -$1,500 means the company is losing $1,500 when producing 30 masks. P'(30) = $200 means profit is increasing by roughly $200 for each additional unit at that production level.
Actual vs. marginal: The actual revenue of the 61st unit is R(61) - R(60). Marginal revenue R'(60) is the approximation. Q30 asks for both: actual = R(61) - R(60) = $2,849, and marginal estimate = R'(60) = $2,880.
Intervals of increase/decrease: f'(x) > 0 means increasing; f'(x) < 0 means decreasing. Read these directly from a sign chart or from a graph of f'(x).
Q14: from the graph, f'(x) > 0 on (-2, 1) union (3, infinity). Answer: C.
Local extrema using the first derivative test: where f' changes sign from + to -, you have a local max; from - to +, a local min.
Second derivative test (Q17): f'(x) = 0 at x = 1 and x = 5, and f''(x) = sqrt(2x - 1) - 2. f''(1) = sqrt(1) - 2 = -1 < 0, so local max at x = 1. f''(5) = sqrt(9) - 2 = 1 > 0, so local min at x = 5. Answer: A.
Concavity: f''(x) > 0 means concave up; f''(x) < 0 means concave down. Q16 asks where f''(x) < 0: read from the graph where the curve bends downward.
Inflection points: where f''(x) = 0 and concavity changes.
Q15: y = 3x^5 - 10x^4. y' = 15x^4 - 40x^3. y'' = 60x^3 - 120x^2 = 60x^2(x - 2). Setting y'' = 0 gives x = 0 and x = 2. Check concavity change: at x = 0, f'' does not change sign (both sides are non-positive), so x = 0 is not an inflection point. At x = 2, f'' changes from negative to positive, so (2, -64) is an inflection point. Answer: C.
Absolute extrema on [a, b] (Q45): evaluate f at all critical numbers in (a, b) and at the endpoints a and b. The largest value is the absolute max, the smallest is the absolute min.
Q45: f(x) = x^4 - 8x^2 - 16 on [-1, 4]. f'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 0, giving x = 0, x = 2, x = -2. Only x = 0 and x = 2 lie in [-1, 4]. f(-1) = 1 - 8 - 16 = -23. f(0) = -16. f(2) = 16 - 32 - 16 = -32. f(4) = 256 - 128 - 16 = 112. Absolute min at (2, -32), absolute max at (4, 112).
Q18: Find two numbers whose difference is 71 and whose product is a minimum. Let x and x - 71 be the numbers. Product P = x(x - 71) = x^2 - 71x. P' = 2x - 71 = 0, so x = 35.5. The other number is 35.5 - 71 = -35.5. Product: -35.5 * 35.5 = -1,260.25. Answer: E (none of the listed options gives the minimum product with the correct pair).
The definite integral from a to b of f(x) dx = F(b) - F(a), where F is any antiderivative of f.
Q19: integral from 2 to 4 of (3x - 2) dx. Antiderivative: (3/2)x^2 - 2x. Evaluate: [(3/2)(16) - 8] - [(3/2)(4) - 4] = [24 - 8] - [6 - 4] = 16 - 2 = 14. Answer: C.
Key antiderivative rules: integral of x^n dx = x^(n+1)/(n+1) + C; integral of 1/x dx = ln|x| + C; integral of e^x dx = e^x + C.
Q20: integral from 12 to 12 of anything dx = 0. The upper and lower limits are the same, so the integral is zero. Answer: C.
Properties (Q21): integral from a to b of f(x) dx = F(b) - F(a) (Fundamental Theorem); integral of [f(x) +/- g(x)] dx = integral of f(x) dx +/- integral of g(x) dx; integral of (1/x) dx = ln|x| + C. Statements II and III from Q21 are true. Statement I is written with a minus sign (F(b) - F(a) appears as F(b) + F(a) in the original, with a correction), so check the exact wording. Answer: B.
Monthly compounding (Q7): A = 800(1 + 0.046/12)^(12*10) = 800(1.003833...)^120 = $1,266.15. Answer: C.
Continuous compounding (Q6): 30,000 = 3,000 * e^(0.037t). Divide by 3,000: 10 = e^(0.037t). Take ln: ln(10) = 0.037t. t = ln(10)/0.037 = approximately 62 years. Answer: B.
Average rate of change (Q4): (3.43 - 2.85) / (2014 - 2007) = 0.58 / 7 = approximately $0.08 per year. Answer: C.
Exponential decay (Q31): 700 grams decays to 200 grams after 50 years. 200 = 700 * e^(50r). Solve: e^(50r) = 2/7, so 50r = ln(2/7), r = ln(2/7)/50 = approximately -0.02506. Rate of decay: r is approximately -0.02506 (or about -2.506% per year). For part (b), after 80 years: Q = 700 * e^(-0.02506 * 80) = 700 * e^(-2.005) = approximately 94 grams.
N(t) = 100t^2 / (t^2 + 40). N'(t) uses the quotient rule: N'(t) = [200t(t^2 + 40) - 100t^2(2t)] / (t^2 + 40)^2 = 8000t / (t^2 + 40)^2.
N(5) = 2500/65 = approximately 38.46 thousand subscribers after 5 months.
N'(5) = 40000 / 65^2 = approximately 9.47 thousand new subscribers per month at t = 5.
Marginal analysis: N(5) + N'(5) gives an estimate of total subscribers after 6 months: approximately 47.93 thousand.
Tangent line: y - f(a) = f'(a)(x - a)
Profit: P(x) = R(x) - C(x)
Revenue: R(x) = p(x) * x (price times quantity)
Marginal cost / revenue / profit: C'(x), R'(x), P'(x)
Average cost: C-bar(x) = C(x) / x
Average profit: P-bar(x) = P(x) / x
Second derivative test: f'(c) = 0 and f''(c) > 0 means local min; f''(c) < 0 means local max
Definite integral (Fundamental Theorem): integral from a to b of f(x) dx = F(b) - F(a)
Power rule for integration: integral of x^n dx = x^(n+1) / (n+1) + C, for n not equal to -1
Integral of 1/x: ln|x| + C
Integral of e^x: e^x + C
Periodic compound interest: A = P(1 + r/n)^(nt)
Continuous compounding: A = Pe^(rt)
Exponential growth/decay: Q(t) = Q0 * e^(rt); r > 0 is growth, r < 0 is decay
Marginal analysis is the core of business calculus. Every pricing, production, and investment decision asks some version of "is one more unit worth it?" The derivative answers that question directly. Compound interest formulas govern savings accounts, loans, and investment growth. Exponential decay models are used in depreciation of assets, radioactive decay in environmental science, and the decline of product demand over time.
Students confuse marginal cost C'(x) with average cost C(x)/x. Marginal cost is the derivative and tells you what the next unit costs. Average cost divides total cost by quantity. They are different quantities with different formulas.
When finding absolute extrema on a closed interval, students often forget to check the endpoints. The absolute max or min can occur at an endpoint, not just at a critical number.
Students mix up the compound interest formula (discrete) with continuous compounding. If the problem says "compounded monthly," use A = P(1 + r/n)^(nt) with n = 12. If it says "compounded continuously," use A = Pe^(rt). Using the wrong formula guarantees the wrong answer.
An inflection point requires a change in concavity, not just f''(x) = 0. Setting f''(x) = 0 gives candidates, but you must verify that f'' changes sign at that point.
⚠️ Tangent line questions appear in both multiple-choice and free-response. Always write the final answer in slope-intercept form (y = mx + b) unless told otherwise.
⚠️ Marginal analysis is tested with interpretation: you need to state in words what P(30) and P'(30) mean, not just compute them.
⚠️ The second derivative test is faster than the first derivative test for classifying critical points. Know both, but reach for the second derivative test first.
⚠️ "Compounded continuously" vs. "compounded monthly" determines which formula to use. Read the problem wording carefully.
⚠️ Absolute extrema on a closed interval: evaluate at critical numbers AND endpoints. Miss either and you miss the answer.
⚠️ The integral from a to a of any function is 0. This is a free mark if you recognise it.
True or False: The tangent line at a point has the same slope as the curve at that point.
True. The slope of the tangent line equals f'(a) at x = a.
Fill in the blank: Profit = ___ minus ___.
Revenue minus Cost.
True or False: If f''(c) > 0 and f'(c) = 0, then x = c is a local maximum.
False. f''(c) > 0 means concave up, which indicates a local minimum.
True or False: The integral from 5 to 5 of (3x^2 + 7x) dx equals 0.
True. When upper and lower limits are the same, the definite integral is always 0.
Fill in the blank: For continuous compounding, the formula is A = ___.
Pe^(rt).
Q: Find the equation of the tangent line to y = 2x^3 - 3x^2 - 12x + 4 at x = 1.
A: y(1) = 2 - 3 - 12 + 4 = -9. y' = 6x^2 - 6x - 12, so y'(1) = -12. Tangent: y + 9 = -12(x - 1), giving y = -12x + 3.
Q: A company's profit function is P(x) = 5x^2 - 100x - 3000. Find P(30) and P'(30) and interpret both.
A: P(30) = 5(900) - 3000 - 3000 = -1500. At 30 units, the company is losing $1,500. P'(x) = 10x - 100, so P'(30) = 200. At 30 units, profit is increasing by approximately $200 per additional unit.
Q: Evaluate the integral from 2 to 4 of (3x - 2) dx.
A: Antiderivative: (3/2)x^2 - 2x. F(4) = 24 - 8 = 16. F(2) = 6 - 4 = 2. Answer: 16 - 2 = 14.
Q: How long will it take $3,000 to grow to $30,000 at 3.7% compounded continuously?
A: 30,000 = 3,000e^(0.037t). Divide: 10 = e^(0.037t). ln(10) = 0.037t. t = 2.3026/0.037 = approximately 62 years.
Q: $800 is invested at 4.6% compounded monthly for 10 years. What is the final amount?
A: A = 800(1 + 0.046/12)^(120) = 800(1.003833)^120 = approximately $1,266.15.
Q: Find all inflection points of y = 3x^5 - 10x^4.
A: y'' = 60x^3 - 120x^2 = 60x^2(x - 2). Candidates: x = 0 and x = 2. At x = 0, y'' does not change sign (stays non-positive), so no inflection point. At x = 2, y'' changes from negative to positive. y(2) = 96 - 160 = -64. Inflection point: (2, -64).
Q: Find two numbers whose difference is 71 and whose product is a minimum.
A: Let the numbers be x and x - 71. P(x) = x^2 - 71x. P'(x) = 2x - 71 = 0, giving x = 35.5 and x - 71 = -35.5. The pair is 35.5 and -35.5.
Tangent lines and marginal analysis rely entirely on differentiation rules from Part 2. If you are shaky on the chain rule or product rule, revisit those before tackling applied problems.
Curve sketching ties together first derivatives (increasing/decreasing), second derivatives (concavity), limits (asymptotes), and continuity into a single comprehensive analysis. It is the capstone skill of the course.
Integration is the inverse of differentiation. Every antiderivative rule mirrors a derivative rule run backwards, so mastery of derivatives makes integration far more intuitive.
tangent line, slope-intercept form, point-slope form, marginal cost, marginal revenue, marginal profit, revenue function, cost function, profit function, average cost, average profit, increasing, decreasing, critical number, local maximum, local minimum, second derivative test, first derivative test, concavity, concave up, concave down, inflection point, absolute extrema, closed interval, definite integral, Fundamental Theorem of Calculus, antiderivative, compound interest, continuous compounding, exponential growth, exponential decay, optimisation, business calculus, MAT101, calculus for business